The following reaction is at equilibrium at 298 K<br>.<br> for the reaction is — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

$K_p = \frac{P_{NOCl}^2}{P_{NO}^2 P_{Cl_2}} = \frac{(0.01)^2}{(0.00001)^2 (0.01)} = \frac{10^{-4}}{10^{-10} \times 10^{-2}} = \frac{10^{-4}}{10^{-12}} = 10^8$. $\Delta G^\circ = -RT \ln K_p = -8.314 \times 298 \times \ln(10^8) = -2477.5 \times 8 \times 2.303 = -45638\text{ J} \approx -45.64\text{ kJ}$.